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Proportion Of Ones

Problem

If we consider 2-digit numbers, there are exactly 18 numbers that contain the digit one:

10, 11, 12, 13, 14, 15, 16, 17, 18, 19,
21, 31, 41, 51, 61, 71, 81, 91

As there are ninety 2-digit numbers, the probability of a 2-digit number containing a one is 18/90 = 1/5.

What proportion of 3-digit numbers contain the digit one?


Solution

There are three mutually exclusive cases for consideration:
1**, (1')1*, and (1')(1')1.

First DigitSecond DigitThird DigitCombinations
10-90-91times10times10 = 100
2-910-98times1times10 = 80
2-92-9 + 018times9times1 = 72
Total
252

There are 9times10times10 = 900 numbers that contain three digits.

Hence the proportion of 3-digit numbers containinig the digit one is 252/900 = 7/25.

Would you expect the proportion of 4-digit numbers containing the digit one to be more or less?
What about n-digit numbers?

Problem ID: 189 (28 Nov 2004)     Difficulty: 2 Star

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